Question

@ganeshie8 problem 28. Please, explain me

Answer 1

\(\sum_{i=1}^n i^3 = \dfrac{1}{4}(n^4 + 2n^3 +n^2)\), and the last term on it is \(n^3\)
Therefore \(\sum_{i=1}^{n-1} = \dfrac{n^2}{4}(n-1)^2\)
What I don't get is that : \(n|(n^2)\) for all n, right? why do we just prove in cases n is odd and n is divisible by 4?

Answer 2

that 4 in the denomiantor...
you need to think about that seriously

Answer 3

if \(n\) is even, then \(n-1\) is odd.
then the bottom \(4\) will kill a factor of \(2\) from the top \(n^2\)

Answer 4

What if we write it as \(n^2 *\dfrac{(n-1)^2}{4}\)

Answer 5

what good is that ?
if \(n\) is even, then definitely \(4\) cannot divide \((n-1)^2\)

Answer 6

divisibility arguments make sense only for integers

Answer 7

oh, got what you mean

Answer 8

when you see an expression like \(n^2 *\dfrac{(n-1)^2}{4}\),
first, you need to worry about whether it will be an integer or not. then you can think of divisibility arguments

Answer 9

Got you, continue, please

Answer 10

In our case, clearly, \(n^2 *\dfrac{(n-1)^2}{4}\) is an integer, why ?

Answer 11

To me, \((\dfrac{(n(n-1)}{2})^2\) and n(n-1) is an even number divided by 2, hence it is an integer.

Answer 12

Thats right, but there is a much much simpler way to see that

Answer 13

what does the expression \((\dfrac{(n(n-1)}{2})^2\) represent exactly ?

it represents sum of cubes of "integers", so it better be an integer, right ?

Answer 14

YYes!!

Answer 15

good, so i think you can try the proof now

Answer 16

Still not pass why 4 instead of 2? As above, if n is odd, 4 not | (n-1)^2, then 4| n^2
But only thing I can get from this is 2 | n

Answer 17

suppose \(n \) is just even and not divisible by \(4\)

Answer 18

we will get

\(\dfrac{n^2}{4}(n-1)^2 = \dfrac{n*n}{4} (n-1)^2 = \text{odd integer}\)

why ?

Answer 19

if that expression is odd, and n is even,
then clearly that expression is not divisible by n

Answer 20

Yes.

Answer 21

\(\dfrac{n^2}{4}(n-1)^2 = \dfrac{n*n}{4} (n-1)^2 \)

if \(n\) is divisible by \(4\), then the bottom \(4\) disturbs only one \(n\) in the top
and we will have the other \(n\) left untouched in the top

Answer 22

ok

Answer 23

maybe consider a quick example

Answer 24

let \(n = 12\) and \(n=10\)

Answer 25

its not hard to see that \(10\nmid \left(\dfrac{10(10-1)}{2}\right)^2\)

Answer 26

because the right hand side is odd

Answer 27

We can go this way
\(\dfrac {n.n (n-1)^2}{4} \)
if n is even and not | by 4, then the expression is not an integer, which is impossible.
Hence if n is even, it must be divisible by 4 to make it true
if n is odd, then n-1 is even, n -1 = 2k , (n-1)^2 = 4k^2 and 4| (n-1)^2 And then n | the whole thing.
Does it make sense?

Answer 28

I lost the net!!!! Can see nothing...... :(

Answer 29

Nope. The expression is "always" an integer. We have argued that earlier

Answer 30

cube of an integer is another integer

sum of integers is another integer